The expected value of a set of outcomes can be found by taking the weighted sum of the outcomes, with the probabilities of the weights:
[tex] E([X])=$\sum_{i=1}^{n}x_{i}p_{i}$ [/tex]
[tex] E(fieldgoal)=(x_{1}p_{1})+(x_{2}p_{2})[/tex]
[tex] E(fieldgoal)=((3pts)(0.90))+((0pts)(0.10))[/tex]
[tex] E(fieldgoal)=2.7pts+0pts[/tex]
[tex] E(fieldgoal)=2.7pts[/tex]
Answer: The expected value of a field goal is 2.7.
Step-by-step explanation:
Since we have given that
Probability of making a field goal for 3 points = 90%
Probability of making a field goal for 7 points = 35%
So, Expected value of a field goal is given by
[tex]E(x)=\sum xp(x)[/tex]
X= 3 points ( make a field goal) and P(3)= 0.90
and
X = 0 points ( make no field goal) and P(7)= 1-0.90=0.10
[tex]E(x)=3\times \dfrac{90}{100}+0\times \dfrac{10}{100}\\\\E(x)=3\times 0.9+0\times 0.10\\\\E(x)=2.7[/tex]
Hence, the expected value of a field goal is 2.7.
